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Chapter 14: Problem 9

The common ion effect for weak acids is to significantly decrease the dissociation of the acid in water. Explain the common ion effect.

Short Answer

Expert verified
The common ion effect is the suppression of weak acid dissociation in water when a solution already contains an ion involved in the equilibrium. According to Le Chatelier's Principle, when a common ion is added, the system shifts the equilibrium to counteract the change, leading to decreased dissociation of the weak acid and a smaller hydrogen ion concentration, ultimately affecting the pH of the solution.

Step by step solution

01

Understand Weak Acid Dissociation

A weak acid is an acid that does not fully dissociate in solution, meaning that not all of its acidic hydrogen atoms are donated to water. The dissociation of a weak acid (HA) can be represented as: \(HA(aq) \rightleftharpoons H^{+}(aq) + A^{-}(aq)\) The double arrow indicates that the reaction is reversible and reaches an equilibrium.
02

Know the Concept of Equilibrium

Equilibrium is reached when the rate of the forward reaction (products forming) is equal to the rate of the reverse reaction (reactants forming). At equilibrium, the concentrations of reactants and products remain constant. For weak acid dissociation, this equilibrium can be represented using the acid-dissociation constant (Ka): \(K_a = \dfrac{[H^{+}][A^{-}]}{[HA]}\) Where [H鈦篯, [A鈦籡, and [HA] represent the equilibrium concentrations of hydronium ions, conjugate base anions, and the weak acid, respectively.
03

Understand Common Ion Effect

The common ion effect occurs when a solution already contains an ion that is involved in the dissociation of a weak acid. This effect influences the equilibrium by suppressing the dissociation of the weak acid.
04

Apply Le Chatelier's Principle to the Common Ion Effect

Le Chatelier's Principle states that if a change is made to a system at equilibrium, the system will shift its equilibrium position to counteract that change. When a common ion is added to a weak acid solution, the system tries to reduce the stress of the added ion by shifting the equilibrium position. In the case of weak acid dissociation: \(HA(aq) + H_2O(l) \rightleftharpoons H^{+}(aq) + A^{-}(aq)\) When a common ion (A鈦�) is added, the system shifts the equilibrium to the left in order to counteract the increased concentration of A鈦� ions. This leads to decreased dissociation of the weak acid and a smaller hydrogen ion concentration, which causes the pH to rise.
05

Conclusion

The common ion effect is the suppression of weak acid dissociation in water when a solution already contains an ion involved in the equilibrium. The equilibrium shifts in response to the added ion, according to Le Chatelier's Principle, leading to a change in the system's pH.

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Most popular questions from this chapter

Which of the following mixtures would result in buffered solutions when 1.0 L of each of the two solutions are mixed? a. \(0.1 M\) KOH and \(0.1 M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.1 M\) KOH and \(0.2 M \mathrm{CH}_{3} \mathrm{NH}_{2}\) c. \(0.2 M\) KOH and \(0.1 M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) d. \(0.1 M \mathrm{KOH}\) and \(0.2 M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\)

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M\) KOH. Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{mL}\) b. \(50.0 \mathrm{mL}\) c. \(100.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100 \(M\) lactic acid (HC \(_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), \(\mathrm{p} K_{\mathrm{a}}=3.86\) is titrated with \(0.100 \mathrm{M}\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL}\) \(20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\) \(26.0 \mathrm{mL}, 28.0 \mathrm{mL},\) and \(30.0 \mathrm{mL}\) of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Consider the following acids and bases: Choose substances from the following list that would be the best choice to prepare a \(\mathrm{pH}=9.0\) buffer solution. a. \(\mathrm{HCO}_{2} \mathrm{H}\) b. HOBr c. \(\mathrm{KHCO}_{2}\) d. \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) \(\mathbf{e} .\left(\mathbf{C}_{2} \mathbf{H}_{5}\right)_{2} \mathrm{NH}\) f. \(\left(C_{2} H_{5}\right)_{2} N H_{2} C l\) g. \(\mathrm{HONH}_{2}\) h. NaOBr
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